# elasticity from inverse demand

I was wondering of my thinking here was right:

Given $$e=\frac{dQ}{dp}*\frac{p}{Q},$$ where $$e$$ is elasticity, $$dQ/dp$$ is first derivative of demand function, $$p$$ is price and $$Q$$ is quantity.

With this expression for $$e$$, could you then state that:

$$\frac{dp}{dQ} = \frac{p}{(e*Q)}$$

where $$dp/dQ$$ expresses first derivative of inverse demand.

Is this correct?

I will denote the demand function by $$Q(p)$$ and the inverse demand function by $$P(q)$$. Then $$\forall q: Q(P(q)) = q$$ so for any $$h > 0$$ and $$q$$ we have \begin{align*} p & := P(q) \\ p_h & := P(q+h) \\ q & = Q(p) \\ q_h & := Q(p_h) = q+h \end{align*} From the definition of derivatives $$\frac{\text{d} P(q)}{\text{d} q} := \lim_{h \to 0} \frac{p_h - p}{q_h - q}.$$ If this is a non zero real number then indeed $$\frac{1}{\lim_{h \to 0} \frac{p_h - p}{q_h - q}} = \lim_{h \to 0} \frac{q_h - q}{p_h - p} = \frac{\text{d} Q(p)}{\text{d} p}.$$
There are some technical considerations. We did not prove that $$\lim_{h \to 0} p_h - p = 0.$$ This is true if $$P$$ is continuous.
We also assumed that at $$q$$ and $$q+h$$ both functions are single valued. If there is a price $$p$$ where $$Q(p) = 0$$ then $$\frac{\text{d} P(q)}{\text{d} q}$$ does not exist.
Yes, for the standard case of a strictly decreasing demand function $$Q(p)$$ and price-elasticity of demand $$\epsilon_p(Q)=Q'(p)\frac{p}{Q(p)}$$ the inverse demand function $$p(Q)$$ exists and by the inverse function theorem $$p'(Q)=\frac{1}{Q'(p)}$$. This gives $$p'(Q)=\frac{p(Q)}{\epsilon_p(Q)Q}$$ wherever the derivatives exist.